What is the integral of e^{y/(x+y)} ?

The integral of e^{y/(x+y)} is ye^{y/(y+x)}+e^{y/(y+x)}x-y\Ei(y/(y+x))+C

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integral of e^{y/(x+y)}

\int e^{\frac{y}{x + y}} d x

y e^{\frac{y}{y + x}} + e^{\frac{y}{y + x}} x - y Ei (\frac{y}{y + x}) + C

Solution steps

\int e^{\frac{y}{x + y}} d x

Apply u-substitution

: \int e^{\frac{y}{u}} d u

= \int e^{\frac{y}{u}} d u

Apply u-substitution

: \int - \frac{y e ^{v}}{v ^{2}} d v

= \int - \frac{y e ^{v}}{v ^{2}} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - y \cdot \int \frac{e ^{v}}{v ^{2}} d v

Apply Integration By Parts

: - \frac{e ^{v}}{v} - \int - \frac{e ^{v}}{v} d v

= - y (- \frac{e ^{v}}{v} - \int - \frac{e ^{v}}{v} d v)

\int - \frac{e ^{v}}{v} d v = - Ei (v)

= - y (- \frac{e ^{v}}{v} - (- Ei (v)))

Substitute back

= - y (- \frac{e ^{\frac{y}{y + x}}}{\frac{y}{y + x}} - (- Ei (\frac{y}{y + x})))

Simplify

- y (- \frac{e ^{\frac{y}{y + x}}}{\frac{y}{y + x}} - (- Ei (\frac{y}{y + x}))) : y e^{\frac{y}{y + x}} + e^{\frac{y}{y + x}} x - y Ei (\frac{y}{y + x})

= y e^{\frac{y}{y + x}} + e^{\frac{y}{y + x}} x - y Ei (\frac{y}{y + x})

Add a constant to the solution

= y e^{\frac{y}{y + x}} + e^{\frac{y}{y + x}} x - y Ei (\frac{y}{y + x}) + C

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What is the integral of e^{y/(x+y)} ?
The integral of e^{y/(x+y)} is ye^{y/(y+x)}+e^{y/(y+x)}x-y\Ei(y/(y+x))+C